Prove ${a_n} = \lim_{n \to \infty } \sum\nolimits_{k \ge 1} {{1 \over {{k^2}}}} $ Converges by using Cauchy's criteria

Question

Prove ${a_n} = \lim\limits_{n \to \infty } \sum\nolimits_{k \ge 1} {{1 \over {{k^2}}}} $ Converges by using Cauchy's criteria.

What I did:
Let $n, m=n+k \in \mathbb{N}$.

$$\left| {{a_m} - {a_n}} \right| = \left| {{a_{n + k}} - {a_n}} \right| = \left| {\sum\limits_{l = n + 1}^{n + k} {{1 \over {{l^2}}}} } \right| \le k \cdot {1 \over {{{(n + 1)}^2}}} \le {k \over n}$$

At this phase I need to choose $N\in\mathbb{N}$ such that $\forall m,n>N: \left| {{a_m} - {a_n}} \right| < \varepsilon$

The problem is that $k\over n$ depends on $k$. How to "correct" this?

Answer 1

Hint: $$ \begin{align} \sum_{k=m+1}^n\frac1{k^2} &\le\sum_{k=m+1}^n\frac1{k(k-1)}\\ &=\sum_{k=m+1}^n\left(\frac1{k-1}-\frac1k\right)\\ &=\frac1m-\frac1n \end{align} $$

Answer 2

Here is a more general approach suppose $\sum a_n$ is a divergent series of positive terms and denote $s_n$ its partial sum.

We shall prove using Cauchy's criterion that $\sum a_n/s_n^2 $ converges. Given $\varepsilon>0$. There is some $n_0$ such that $s_{n_0}> 1/ \varepsilon$. Then for $q> p\ge n_0\,$ we have

\begin{align}\sum_{k=p+1}^q\frac{a_n}{s_n^2}<\sum_{k=p+1}^q\frac{a_n}{s_n(s_{n-1})} =\sum_{k=p+1}^q\frac{s_n-s_{n-1}}{s_n(s_{n-1})}\\=\sum_{k=p+1}^q\frac{1}{s_{n-1}}-\frac{1}{s_{n}}=\frac{1}{s_p}-\frac{1}{s_q}\\<\frac{1}{s_p}\le\frac{1}{s_{n_0}}<\varepsilon\end{align}

Note that if $a_n=1$, the above proposition give us that $\sum1/n^2$ convergence using Cauchy's criterion.